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selfAdjoint
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Why don't you start by showing why y[3x^2 +3xy + y^2] can't be a cube, and then we'll go on from there.
Originally posted by selfAdjoint
Why don't you start by showing why y[3x^2 +3xy + y^2] can't be a cube, and then we'll go on from there.
Originally posted by Russell E. Rierson
[x+y]^2 - x^2 = 2xy + y^2
3^2 = 2*4*1 + 1^2
5^2 = 2*12*1 + 1^2
7^2 = 2*24*1 + 1^2
[x+y]^3 - x^3 = 3yx^2 + 3xy^2 + y^3
y[3x^2 +3xy + y^2]
cannot be a cube.
y[2x+y] is a square.
y[3x^2 +3xy + y^2] is not a cube
y[4x^3 + 6yx^2 + 4xy^2 + y^3] is not a 4th power
Originally posted by matt grime
And that is never a cube for what reason? And the other infinite set of cases for n=3 are also false because?
I don't remember a proof that 1 more than 6 times the sum of the first N numbers is a cube, so fill in the blanks for me.
Originally posted by matt grime
The universal set? Representable? Do you practise at this or does it come naturally?
digiflux said:Princeton has gotten tremendous publicity from Wiles and his bogus "proof". That means $ pour in. They ARE protecting their crybaby, Wiles.